24.kgci z -o23ino acis 76, were, producea, what the percent veld? 12, wrte the b
ID: 588165 • Letter: 2
Question
24.kgci z -o23ino acis 76, were, producea, what the percent veld? 12, wrte the balaricnd cornpiere ionic and net lonic equations far t the followri,eocron. 13. Consider the following neutraiiestion reaction:uFCfan aNOHfaa) A 30.00 equivalence point is reached in 3 trials when 26.39 concerntration of the H.PO4 solution? t sampie of an unknown concentratian of acid is titrated with a 0.1007 M NaOH solution. The mL, 24.85 TL and 25.01 ml of NaOH are added. What is the What is the pH of the HsPO4 solution?Explanation / Answer
13a) The balanced chemical equation for the reaction is given. As per the stoichiometric equation,
1 mole H3PO4 = 3 moles NaOH
Consider the millimoles of NaOH used in the three trials.
Trial 1: Millimoles NaOH = (volume of NaOH in mL)*(concentration of NaOH in M) = (26.38 mL)*(0.1007 M) = 2.656466 mmole.
Trial 2: Millimoles NaOH = (volume of NaOH in mL)*(concentration of NaOH in M) = (24.85 mL)*(0.1007 M) = 2.502395 mmole.
Trial 3: Millimoles NaOH = (volume of NaOH in mL)*(concentration of NaOH in M) = (25.01 mL)*(0.1007 M) = 2.518507 mmole.
Average millimoles of NaOH = 1/3*(2.656466 + 2.502395 + 2.518507) mmole = 2.559122667 mmole.
As per the stoichiometric equation,
2.559122667 mmole NaOH = (2.559122667 mmole NaOH)*(1 mole H3PO4/3 moles NaOH) = 0.853040888 mmole H3PO4.
30.00 mL of H3PO4 was taken; therefore, the concentration (molarity) of H3PO4 is (millimoles of H3PO4)/(volume of solution) = (0.853040888 mmole)/(30.00 mL) = 0.028434696 mmole/mL 0.0284 M (ans).
b) Write down the ionization of H3PO4 as below.
H3PO4 (aq) -------> H+ (aq) + H2PO4- (aq)
The acid dissociation constant is given as
Ka = [H+][H2PO4-]/[H3PO4] = 6.9*10-3
====> 6.9*10-3 = (x).(x)/(0.0284 – x) (note the 1:1 nature of dissociation).
Assume x << 0.0284 M; this is true because the acid dissociation constant is small. Therefore,
6.9*10-3 = x2/(0.0284)
====> x2 = 1.9596*10-4
====> x = 0.013998
Therefore, [H+] = 0.013998 M and pH = -log [H+] = -log (0.013998) = 1.85393 1.85 (ans).
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